So I'm working with a skilsaw, I don't have a single extension cord long enough so I join two together. One is thicker than the other. The guy I'm working with, Jimmy, has been doing construction longer than I have been alive. He tells me, "Make sure you put the thicker cord closer to the outlet and the thinner cord towards the saw." I ask him "does it matter?" "Of course. Closer to the outlet has more power so it needs the thicker cord. Out towards the saw there's less power so it doesn't need as thick a cord." Jimmy is a smart guy, a good friend, and we enjoy taking the time to talk these things through.

He explains, "When we had the saw hooked just to the shorter thicker cord you could hear and see the saw cut faster. There's power loss along any cord thick or thin, so the saw slows down the longer the cord. Right?" "Of course, Jimmy." "Well closer to the outlet, there's more power available. If we put the thin cord there it could burn up because the thicker cord and saw might draw more power than the thin cord could handle."

I tell him "We know that amps are equal all across a closed loop, so it shouldn't matter because the amps are the same anywhere along either cord." Jimmy gets a gleam in his eye, "And watts are amps times volts. So if there's less power at the saw, that means less watts, and if amps are the same anywhere along the cord, that means volts have gone down with the longer cord. And lower voltage doesn't need as thick of a cord. So the thin cord goes at the end."

Jimmy stubs out his cigarette, and finishes the discussion "So either you're right and it doesn't matter, or I'm right and it does. So let's do it my way just to be safe and get back to work. Break's over."

Is Jimmy right? If not, can you explain it in layman's terms better than he did? I'm never going to change his mind of course, but I'd like to know all the same.

Each cord has two wires in it (I'm assuming), each of which has a bit of resistance, so can be treated as a resistor, which I'll call Ri. So, the circuit you get is: from one of the holes in the outlet; along one of the wires of the first cord, R1; along one of the wires of the second cord, R2; through the skilsaw; along the other wire of the second cord, R3; along the other wire of the first cord, R4; back to the other hole in the outlet; and, finally, through whatever the electricity supply is back to the first hole in the outlet.

So, between the outlet and the skilsaw, you have two little series networks of resistors: R1 + R2 and R3 + R4. The resistances are added together, because the total resistance of two resistors in series is simply the sum of their resistances. If you swap the cords around, you're swapping the resistors in each little network: R2 + R1 and R4 + R3. But Ra + Rb = Rb + Ra, because x+y = y+x. Addition is said to be commutative: it's the same whichever way round you add the terms.

Since swapping the cords round doesn't change the resulting resistances, the two circuits must be equivalent. Whatever voltage drops you get across R1 + R2 and R3 + R4 when the cords are the first way round, you must also get across R2 + R1 and R4 + R3 when the cords are the other way round. And nothing else is changing. So, you'll have the same voltage drop across the skilsaw, the same voltage across the outlet, and the same current around the circuit.

The voltage drop and current through R1 will be the same regardless of which way round the cords are. The same goes for the other three wires in the cords. There is no advantage from having the thicker cord nearer the outlet. It's the same current going through the cords either way, the same voltage drops along the wires, and the same power losses along them.

So, it boils down to series resistors and basic circuit theory.

Perhaps the thing to emphasize is that the power loss along a cord doesn't depend on the overall voltage drop as seen from one end of it, but depends on the two voltage drops along each wire in it. If you've got the thicker cord plugged into the outlet, yes, you will have a larger voltage drop across that end of it, but you'll also have a larger voltage drop across the other end of it, since it's got both the other cord and the skilsaw across it instead of just the skilsaw. The voltage drops along the wires of the thicker cord will still be the same as when the cords are the other way around. And it's the voltage drop along each wire that matters, since it's along each wire that there will be a bit of power loss.

Does that explanation work so far?

I could draw some diagrams later to illustrate this, but I'll leave it at this for now.

Jimmy is wrong, FancyHat is right. I just want to boil down FancyHat's answer a little with a crude diagram:

Jimmy's mistake is that he's thinking about V_big. However, V_small1 and V_small2 are what control the power dissipation in the cord, and those voltages are entirely determined by the resistance of the cord, and the current, neither of which depend on what order you put the cords in.

Perhaps I’m not thinking this through clearly, but does it make a difference that the current is AC rather than DC? As in, whichever cord is proximal to the outlet will experience a greater voltage fluctuation, whereas the distal cord will see the maximal voltage already somewhat reduced. I would think it desirable to lessen the strain on the thinner cord.

Cervisiae Amatorem wrote:"When we had the saw hooked just to the shorter thicker cord you could hear and see the saw cut faster."

Look if you think are you too smart for Jimmy just because you know Ohm's laws, you should know that you are not too smart for observation.The thin cord is probably so shitty that you get some nonlinearity, probably due to the heat.

LE4dGOLEM: What's a Doug?Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Qaanol wrote:Perhaps I’m not thinking this through clearly, but does it make a difference that the current is AC rather than DC? As in, whichever cord is proximal to the outlet will experience a greater voltage fluctuation, whereas the distal cord will see the maximal voltage already somewhat reduced. I would think it desirable to lessen the strain on the thinner cord.

I don't think the cord cares about the voltage difference between the two wires in it, just about the difference between the ends of the individual wires (unless the cord is damaged and there is some current flowing directly from one to the other). Also the fluctuation in the voltage along the wires should be proportional to the voltage along the wires, which should be the same in both cases.

doogly wrote:Look if you think are you too smart for Jimmy just because you know Ohm's laws, you should know that you are not too smart for observation.The thin cord is probably so shitty that you get some nonlinearity, probably due to the heat.

Absolutely, nobody is too smart for observation. They should try it both ways and see if there's a difference (letting the system cool down between tests). With regard to temperature and non-linearity though, I'm aware that resistance increases as temperature goes up. However, no matter which order you put the cords in, I think the system will arrive at the same steady state. The thin cord should come up to the same temperature and same resistance in both cases. If you're instead talking about the resistance of the cord changing as a function of voltage due to some temperature induced effect, or temperature causing significant inductance or capacitance, I'm not sure what you're talking about but I don't know everything and would be happy to learn.

Note that I'm not saying the cords are equally good - just that if you have a good cord and a bad cord and must put them in series, the order doesn't matter.

An extension cord acts as a capacitor in parallel with the load. So the argument that all amps go through the entire circuit is not correct, the latter part of the cables sees less current. I am too lazy to calculate if this is significant enough to feel as heat.

I was always taught not to connect a thicker cable behind a thinner cable. Not because of heating, but because people will use the last cable as indication of the whole chain. If you have

30 amp load64 amp cable32 amp cableMains

Then you can just wait until someone with a 45 amp load will see the 64amp cable and connect their machine to it.

Jimmy is completely right, except for that he thinks that the higher voltage requires a thicker wire. Wire thickness depends on current, not voltage. This also renders his conclusion wrong.

Spoiler:

Two additional points: 1. Now a higher voltage does require a thicker insulation, to prevent a short from one wire in the cable to the other. However, both cables are rated for mains voltage so this doesn't matter. Any voltage below the rated voltage is OK.2. Don't use many extension cords on a cheap saw. As others have posted and Jimmy indicated: each extension cord causes a voltage drop. All saws will get damaged if you use them with a too low voltage, cheap saws are less lenient in this than expensive ones.

Mikeski wrote:A "What If" update is never late. Nor is it early. It is posted precisely when it should be.

I just measured the capacitance of a 6 foot computer power cable at 0.23nF. If we assume that an extension cord has similar properties, a 100ft cord might have a capacitance of about 3.8nF. Perhaps it would be measurable, but probably not significant.

scarecrovv wrote:I just measured the capacitance of a 6 foot computer power cable at 0.23nF. If we assume that an extension cord has similar properties, a 100ft cord might have a capacitance of about 3.8nF. Perhaps it would be measurable, but probably not significant.

That would give a reactance of about 700 kΩ at 60 Hz. With a voltage of 110 V AC, the current would be about 160 µA. I don't think it's going to be a problem, even with a heavy, capacitive load on the end.

If the final load is purely resistive, then this extra 160 µA is 𝜋/2 rad out of phase with the load current itself. We then get the net current by using Pythagoras' theorem. So, if the load is drawing 30 A, and is purely resistive, and we've got 3.8 nF in parallel with it and drawing an extra 160 µA, then the net current drawn by that RC parallel network is, uh, so slightly above 30 A that my calculator isn't saying what the extra bit is, even when I subtract 30 A.

With 30 A drawn by a capacitive load instead, we've got 30.00016 A drawn instead of 30 A. Since P=I2/R, the amount of power wasted along the first cable, as a result of the second cable being put between it and the load, is increased by a factor of 1.0000107.

Yeah, I did the calculation afterwards... capacitance only becomes an issue for long range power transmission lines. Much higher voltage, much longer cables. Underground cables suffer from it.

You also encounter the problem with electronics. Some IC with a switch inside draws a rapidly varying current without a decoupling cap nearby, so the PCB lines are effectively transmitting MHz-range alternating current. Sawtoothed even for higher harmonics.Then a 'current in = current out' rule becomes problematic.

But extension cords are safe... I'd go with precaution as background for Jimmy's qualm. The last cable should be the weakest link in the chain, so people can see how much amps the cable chain can handle.

Zamfir wrote:But extension cords are safe... I'd go with precaution as background for Jimmy's qualm. The last cable should be the weakest link in the chain, so people can see how much amps the cable chain can handle.

The reverse argument could be made as well...the last cable should be the one most able to take the physical abuse involved in being connected to the saw, as it's getting twisted and pulled around, stepped on, and possibly having chunks of wood or even a saw dropped on it. The capacity of the power supply should be checked before plugging it in any power-hungry device. If you can't ensure that people will respect the limitation, you should probably just go buy a heavier cord.

Zamfir wrote:An extension cord acts as a capacitor in parallel with the load. So the argument that all amps go through the entire circuit is not correct, the latter part of the cables sees less current. I am too lazy to calculate if this is significant enough to feel as heat.

I was always taught not to connect a thicker cable behind a thinner cable. Not because of heating, but because people will use the last cable as indication of the whole chain. If you have

30 amp load64 amp cable32 amp cableMains

Then you can just wait until someone with a 45 amp load will see the 64amp cable and connect their machine to it.

This. Also, the thicker cords are generally heavier, and more of a pain if you're on a ladder, or otherwise dealing with a lot of weight/bulk in cord in awkward spaces. Safety and convenience indicates that the thinner cord should be the one at the end of the chain, attached to the load.

As a bonus, the thinner cable is probably cheaper to replace if you wreck it.

Cervisiae Amatorem wrote:He explains, "When we had the saw hooked just to the shorter thicker cord you could hear and see the saw cut faster. There's power loss along any cord thick or thin, so the saw slows down the longer the cord. Right?" "Of course, Jimmy." "Well closer to the outlet, there's more power available. If we put the thin cord there it could burn up because the thicker cord and saw might draw more power than the thin cord could handle."

If Jimmy's concerned that the thinner cord isn't enough to safely handle the current drawn by the skilsaw and thicker cord, he shouldn't be using the thinner cord at all with the skilsaw.

Do these cords have fuses? Where I am, with domestic mains extension cords, plugs and sockets, it's normal for the plugs of cords to have fuses inside them. If they're the right fuses for the cords, they should blow if the cords are overloaded. If using just one cord isn't going to overload it, then sticking a thicker, higher-current-rated cord after it isn't going to overload it, either.

Do these cords have fuses? Where I am, with domestic mains extension cords, plugs and sockets, it's normal for the plugs of cords to have fuses inside them. If they're the right fuses for the cords, they should blow if the cords are overloaded. If using just one cord isn't going to overload it, then sticking a thicker, higher-current-rated cord after it isn't going to overload it, either.

My experience was with heavy, industrial cabling for 3-phase mains. These cables have different plugs depending on amperage, you need an interconnect between them. I have never seen a fuse on those cables, though I have seen them on multi-way splitters (that you are only supposed to use at the very end, directly connected to loads, but you know how it goes).

Industrial systems tend to be slightly less protected against user error than consumer grade. The users are not assumed to be experts, but at least professionals who had some training before use. If they consciously want to use a 32 amp cable for a 40 amp load, so be it. Perhaps it's cold outside, and they can feel how hot the cable is getting in the process.

Rules about putting the cables in decreasing order are to prevent unconscious mistakes by third parties. Cjameshuff wants people to check all the way to the power source, but that might be hundreds of metres away, with corners and doors in between. If people see a 64A cable, they should be free to assume that it is at least 64A all the way. And it's up to you to make it so, by placing the cables in decreasing order.

FancyHat wrote:Do these cords have fuses? Where I am, with domestic mains extension cords, plugs and sockets, it's normal for the plugs of cords to have fuses inside them. If they're the right fuses for the cords, they should blow if the cords are overloaded. If using just one cord isn't going to overload it, then sticking a thicker, higher-current-rated cord after it isn't going to overload it, either.

Here in the Netherlands cables usually don't have fuses. Neither do wall sockets (but those are supposed to be 2,5 mm2 with a 16A fuse (=safe) anyway).The only cables with fuses are sort of special items like lightning protected and conditioned power ones.

Mikeski wrote:A "What If" update is never late. Nor is it early. It is posted precisely when it should be.

FancyHat, do you live in the UK? The UK is a bit of an outlier when it comes to household fuses.

The UK's standards allow "ring circuits" in the walls, intended to supply a fairly large number of wall sockets per single circuit. Therefore, the mains fuse of a circuit can be rated for a higher amperage (32A) than the cabling of individual appliances.

In case of a short in an appliance, the mains will deliver that full 32A which is too much for the cabling. To cover this case, the appliances and extension cords have a fuse in their plug, rated for the current their cable can handle.

In most places, the mains are fused at 16A or 20A. And the cables of appliances are expected to survive such currents, at least for a while.

For the people in Europe, the UK, Australia, etc, talking about fuses and current ratings of cables: bear in mind that Jimmy is (probably) in the USA, where the residential voltage is only 110V, roughly half the voltage supplied in most countries outside of the American region (and Japan & North Korea), so his fuses and cables need to cope with roughly double the current for the same power consumed (compared to a country that uses 220V+). I guess it's also worth mentioning that the cables get hotter due to the quadrupled resistive losses.